由同余激发的若干关于 $1/\pi$ 的新级数
数论
2023-02-23 v4 组合数学
摘要
本文推导出一组六个关于 1 / π 1/\pi 1/ π ∑ n = 0 ∞ 41673840 n + 4777111 5780 n W n ( 1444 1445 ) = 147758475 95 π \sum_{n=0}^\infty\frac{41673840n+4777111}{5780^n}W_n\left(\frac{1444}{1445}\right) =\frac{147758475}{\sqrt{95}\,\pi} n = 0 ∑ ∞ 578 0 n 41673840 n + 4777111 W n ( 1445 1444 ) = 95 π 147758475 W n ( x ) = ∑ k = 0 n ( n k ) ( n + k k ) ( 2 k k ) ( 2 ( n − k ) n − k ) x k W_n(x)=\sum_{k=0}^n\binom nk\binom{n+k}k\binom{2k}k\binom{2(n-k)}{n-k}x^k W n ( x ) = ∑ k = 0 n ( k n ) ( k n + k ) ( k 2 k ) ( n − k 2 ( n − k ) ) x k ∑ n = 0 ∞ a n + b m n ∑ k = 0 n ( n k ) 4 \sum_{n=0}^\infty\frac{an+b}{m^n}\sum_{k=0}^n\binom nk^4 n = 0 ∑ ∞ m n an + b k = 0 ∑ n ( k n ) 4 1 / π 1/\pi 1/ π ∑ k = 0 ∞ 4290 k + 367 3136 k ( 2 k k ) T k ( 14 , 1 ) T k ( 17 , 16 ) = 5390 π , \sum_{k=0}^\infty\frac{4290k+367}{3136^k}\binom{2k}kT_k(14,1)T_k(17,16)=\frac{5390}{\pi}, k = 0 ∑ ∞ 313 6 k 4290 k + 367 ( k 2 k ) T k ( 14 , 1 ) T k ( 17 , 16 ) = π 5390 , T k ( b , c ) T_k(b,c) T k ( b , c ) ( x 2 + b x + c ) k (x^2+bx+c)^k ( x 2 + b x + c ) k x k x^k x k
引用
@article{arxiv.2009.04379,
title = {Some new series for $1/\pi$ motivated by congruences},
author = {Zhi-Wei Sun},
journal= {arXiv preprint arXiv:2009.04379},
year = {2023}
}
备注
20 pages.Accepted version for publication in Colloq. Math